LeetCode 303. Range Sum Query - Immutable

Posted 2020-06-13

求数组nums[i,j]的和

思路：另开一sum数组，sum[i]为nums[0,i]的和，所以nums[i,j] = sum[j] - sum[i-1]

 1 class NumArray {
 2 public:
 3     vector<int> sum;
 4     NumArray(vector<int> &nums) {
 5         sum.resize(nums.size(), 0);
 6         sum[0] = nums[0];
 7         int len = nums.size();
 8         for(int i=1; i<len; ++i)
 9             sum[i] = sum[i-1] + nums[i];
10     }
11 
12     int sumRange(int i, int j) {
13         if(sum.size()==0) return 0;
14         if(i == 0) return sum[j];
15         return sum[j] - sum[i-1];
16     }
17 };

 

但是输入为[]时出现Runtime Error，将sum数组的长度+1，AC。

 1 class NumArray {
 2     vector<int> sums;
 3 public:
 4     NumArray(vector<int> &nums) {
 5         sums.resize(nums.size()+1, 0);
 6         for(int i=1; i<=nums.size(); i++){
 7             sums[i]=sums[i-1]+nums[i-1];
 8         }
 9     }
10 
11     int sumRange(int i, int j) {
12         return sums[j+1]-sums[i];
13     }
14 };